Download - Standard Reduction Potentials
Standard Reduction Potentials
Electrochem III
Balancing Oxidation-Reduction Equations
Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.
Balancing Oxidation-Reduction EquationsThis involves
treating the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction.
The Half-Reaction Method
1.Assign oxidation numbers to determine what is oxidized and what is reduced.
2.Write the oxidation and reduction half-reactions.
The Half-Reaction Method3.Balance each half-
reaction.
• Balance elements other than H and O.
• Balance O by adding H2O.
• Balance H by adding H+.
• Balance charge by adding electrons.
4.Multiply the half-reactions by integers so that the electrons gained and lost are the same.
The Half-Reaction Method
5.Add the half-reactions, subtracting things that appear on both sides.
6.Make sure the equation is balanced according to mass.
7.Make sure the equation is balanced according to charge.
The Half-Reaction Method
Consider the reaction between MnO4− and
C2O42− :
MnO4−
(aq) + C2O42−
(aq) Mn2+ (aq) + CO2 (aq)
The Half-Reaction Method•First, we assign oxidation numbers.
•Manganese is oxidized.
•Carbon is reduced.
MnO4− + C2O4
2- Mn2+ + CO2
+7 +3 +4+2
The Half-Reaction MethodC2O4
2− CO2
To balance the carbon, we add a coefficient of 2:
C2O42− 2 CO2
The Half-Reaction MethodC2O4
2− 2 CO2
The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.
C2O42− 2 CO2 + 2 e−
The Half-Reaction MethodMnO4
− Mn2+
The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.
MnO4− Mn2+ + 4 H2O
The Half-Reaction MethodMnO4
− Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+ to the left side.
8 H+ + MnO4− Mn2+ + 4 H2O
The Half-Reaction Method8 H+ + MnO4
− Mn2+ + 4 H2O
To balance the charge, we add 5 e− to the left side.
5 e− + 8 H+ + MnO4−
Mn2+ + 4 H2O
The Half-Reaction MethodNow we evaluate the two half-reactions
together:
C2O42− 2 CO2 + 2 e−
5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O
To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.
The Half-Reaction Method
5 C2O42− 10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8 H2O
When we add these together, we get:
10 e− + 16 H+ + 2 MnO4− + 5 C2O4
2−
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The Half-Reaction Method10 e− + 16 H+ + 2 MnO4
− + 5 C2O42−
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the electrons. Subtracting them, we are left with:
16 H+ + 2 MnO4− + 5 C2O4
2−
2 Mn2++ 8 H2O + 10 CO2
Half Reactions in Basic Solutions
•If a reaction occurs in basic solution, one can balance it as if it occurred in acid.
•Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.
•If this produces water on both sides, you might have to subtract water from each side.
Standard Cell Potentials•A galvanic cell is always a redox reaction that has been separated
into 2 half reactions.
•Each half reaction is assigned a potential from which the cell potential can be derived.
•For example:
•Redox rxn: 2H+(aq) + Zn(s) Zn2+ (aq) + H2 (g)
•The Cell Potential is 0.76 V
Anode CathodeZn (s) Zn2+ (aq) +2e- 2H+(aq) + 2e- H2 (g)
Standard Hydrogen ElectrodeDefinition:When the platinum cathode is this half reaction has
a concentration of 1 M [H+] with H2 gas at 1 atm.
We can measure Ecell of the total cell at 0.76 V, but there is no way to measure each half reaction directly.
To deal with this, the standard hydrogen electrode has been assigned a cell potential of 0.0V.
Since the Standard Hydrogen Electrode is assigned 0.0V and the total Ecell = 0.76V,
then Zn(s) Zn2+(aq) +2e- has a potential of 0.76 V
Why???Because E°cell= E°cathode +E°Anode
E°cell= E°H+H2+ E°Zn Zn
2+
The ° indicates standard states are employed with 2H+(aq) + 2e- H2 (g) as 0.0V
We can assign values to all other half reactions.
For example:
Zn (s) + Cu 2+ Zn 2+ + Cu (s)
What are the half reactions?
Anode Zn(s) Zn2+ (aq) + 2e- LEO
Cathode Cu (aq) + 2e- Cu (s) GER
Zn (s) + Cu 2+ Zn 2+ + Cu (s)
E°cell= E°cathode +E°Anode
The E°cell is measured at 1.10 V,
And E°Anode is determined that Zn(s) Zn2+
(aq) + 2e- is 0.76V,Then E°cathode Cu (aq) + 2e- Cu (s) must be
0.34 V
Because 1.10V = 0.76 V + 0.34 Vmeasured known calculated
The value of the standard hydrogen electrode.
The value of the standard hydrogen electrode of 0.0V is accepted by the scientific community.
Standard Reduction PotentialsWhen the cell potentials of half reactions are
given, they are written as:Standard Reduction Potentials: All half
reactions have solutions at 1 M and gases at 1 atm.
Zn2+ (aq) + 2e- Zn (s)Cu+ (aq) + e- Cu (s)2H+(aq) + 2e- H2 (g)
They are all written as a gain of electrons, reduction. GER
Combining 2 half reactions to obtain a balanced oxidation-reduction reaction require 2 manipulations:
1.One of the reactions must be reversed, which means the sign of its E° must be reversed.
2.Since the number of electrons lost by the anode must equal the number of electrons gained by the cathode. The half reactions are multiplied to find the lowest common value. But the value of the E° does not change!!
Note: Standard reduction potentials are intensive properties. It does not matter how many times the reaction occurs, only that it occurs.
Consider a Galvanic CellsBased on the unbalanced ReactionFe 3+ (aq) + Cu (s) Cu 2+ (aq) + Fe 2+
(aq)The two half reactions are as
standard reduction potentials:Fe 3+ (aq) + 1e- Fe 2+ (aq)
E° = 0.77VCu 2+ (aq) + 2e- Cu (s)
E° = 0.34V
E°cell continuedTo balance the cell reaction and
calculate the E° equation number 2 must be reversed and the sign of the E° must be reversed.
Cu (s) Cu 2+ (aq) + 2e- E°= -0.34V
And one must double the first equation to balance the electrons.
2Fe 3+(aq) + 2e- 2Fe 2+(aq) E°=0.77V
NOTE: The value of E°is not multiplied
Sum up the two half reactionsCu (s) Cu 2+ (aq) + 2e- E°= -
0.34V2Fe 3+(aq) + 2e- 2Fe 2+(aq)
E°=0.77V
2Fe3+(aq)+Cu(s) 2Fe2+(aq)+Cu2+
(aq) E°=0.43V
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